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Problem 914

AMC 12 late, AIME early

Algebra Difficulty 4.7 Prove it Saudi Arabian Mathematical Competitions · Saudi Arabia

Let $a>0$ . If the system
$\left\{\begin{array}{l} a^{x}+a^{y}+a^{z}=14-a \\ x+y+z=1 \end{array}\right.$
has a solution in real numbers, prove that $a \leq 8$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Assume, by contradiction, that $a>8$ .
Then $14-a<6$ . Applying AM-GM inequality, we get
$6>14-a=a^{x}+a^{y}+a^{z} \geq 3 a^{\frac{x+y+z}{3}}>3 \cdot 8^{\frac{1}{3}}=3 \cdot 8^{\frac{1}{3}}=6,$
a contradiction.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.